Problem: You are given an integer n from 1 to 10^{18} without leading zeroes.

In one move you can swap any two adjacent digits in the given number in such a way that the resulting number will not contain leading zeroes. In other words, after each move the number you have cannot contain any leading zeroes.

What is the minimum number of moves you have to make to obtain a number that is divisible by 25? Print -1 if it is impossible to obtain a number that is divisible by 25.

Input

The first line contains an integer n (1 ≤ n ≤ 10^{18}). It is guaranteed that the first (left) digit of the number n is not a zero.

Output

If it is impossible to obtain a number that is divisible by 25, print -1. Otherwise print the minimum number of moves required to obtain such number.

Note that you can swap only adjacent digits in the given number.

Examples

Input

5071


Output

4


Input

705


Output

1


Input

1241367


Output

-1

Note

In the first example one of the possible sequences of moves is 5071 → 5701 → 7501 → 7510 → 7150.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int MOD = (int)1e9 + 7;
const int FFTMOD = 1007681537;
const int INF = (int)1e9;
const long long LINF = (long long)1e18;
const long double PI = acos((long double)-1);
const long double EPS = 1e-9;
inline long long gcd(long long a, long long b) {
  long long r;
  while (b) {
    r = a % b;
    a = b;
    b = r;
  }
  return a;
}
inline long long lcm(long long a, long long b) { return a / gcd(a, b) * b; }
inline long long fpow(long long n, long long k, int p = MOD) {
  long long r = 1;
  for (; k; k >>= 1) {
    if (k & 1) r = r * n % p;
    n = n * n % p;
  }
  return r;
}
template <class T>
inline int chkmin(T& a, const T& val) {
  return val < a ? a = val, 1 : 0;
}
template <class T>
inline int chkmax(T& a, const T& val) {
  return a < val ? a = val, 1 : 0;
}
inline long long isqrt(long long k) {
  long long r = sqrt(k) + 1;
  while (r * r > k) r--;
  return r;
}
inline long long icbrt(long long k) {
  long long r = cbrt(k) + 1;
  while (r * r * r > k) r--;
  return r;
}
inline void addmod(int& a, int val, int p = MOD) {
  if ((a = (a + val)) >= p) a -= p;
}
inline void submod(int& a, int val, int p = MOD) {
  if ((a = (a - val)) < 0) a += p;
}
inline int mult(int a, int b, int p = MOD) { return (long long)a * b % p; }
inline int inv(int a, int p = MOD) { return fpow(a, p - 2, p); }
inline int sign(long double x) { return x < -EPS ? -1 : x > +EPS; }
inline int sign(long double x, long double y) { return sign(x - y); }
const int maxn = 2e5 + 5;
int n;
long long x[maxn];
long long toInt(string s) {
  long long res = 0;
  for (int i = (0); i < (int((s).size())); ++i) {
    res = res * 10 + s[i] - '0';
  }
  return res;
}
void volamtruyenkyii() {
  if (fopen("in.txt", "r")) {
    freopen("in.txt", "r", stdin);
  }
  string s;
  cin >> s;
  int n = int((s).size());
  if (n == 1) {
    cout << -1 << "\n";
    return;
  }
  if (n == 2) {
    long long k = toInt(s);
    if (k % 25 == 0) {
      cout << 0 << "\n";
      return;
    }
    swap(s[0], s[1]);
    if (s[0] != '0' && toInt(s) % 25 == 0) {
      cout << 1 << "\n";
      return;
    }
    cout << -1 << "\n";
    return;
  }
  int res = INF;
  for (int i = (0); i < (n); ++i)
    for (int j = (i + 1); j < (n); ++j) {
      int a = s[i] - '0';
      int b = s[j] - '0';
      if ((a * 10 + b) % 25 == 0) {
        int tot = (n - j - 1) + (n - i - 2);
        string t;
        for (int k = (0); k < (n); ++k)
          if (k != i && k != j) {
            t += s[k];
          }
        if (toInt(t) == 0) {
          continue;
        }
        while (t[0] == '0') {
          for (int i = (0); i < (int((t).size()) - 1); ++i)
            if (t[i + 1] != '0') {
              tot++;
              swap(t[i], t[i + 1]);
              break;
            }
        }
        chkmin(res, tot);
      }
    }
  for (int i = (0); i < (n); ++i)
    for (int j = (i + 1); j < (n); ++j) {
      int a = s[i] - '0';
      int b = s[j] - '0';
      if ((b * 10 + a) % 25 == 0) {
        int tot = (n - j - 1) + (n - i - 2) + 1;
        string t;
        for (int k = (0); k < (n); ++k)
          if (k != i && k != j) {
            t += s[k];
          }
        if (toInt(t) == 0) {
          continue;
        }
        while (t[0] == '0') {
          for (int i = (0); i < (int((t).size()) - 1); ++i)
            if (t[i + 1] != '0') {
              tot++;
              swap(t[i], t[i + 1]);
              break;
            }
        }
        chkmin(res, tot);
      }
    }
  if (res == INF) res = -1;
  cout << res << "\n";
}
int main(int argc, char* argv[]) {
  ios_base::sync_with_stdio(0), cin.tie(0);
  if (argc > 1) {
    assert(freopen(argv[1], "r", stdin));
  }
  if (argc > 2) {
    assert(freopen(argv[2], "wb", stdout));
  }
  volamtruyenkyii();
  cerr << "\nTime elapsed: " << 1000 * clock() / CLOCKS_PER_SEC << "ms\n";
  return 0;
}